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A324183
a(n) = d(A163511(n)), where d(n) is A000005, the number of divisors of n.
6
1, 2, 3, 2, 4, 3, 4, 2, 5, 4, 6, 3, 6, 4, 4, 2, 6, 5, 8, 4, 9, 6, 6, 3, 8, 6, 8, 4, 6, 4, 4, 2, 7, 6, 10, 5, 12, 8, 8, 4, 12, 9, 12, 6, 9, 6, 6, 3, 10, 8, 12, 6, 12, 8, 8, 4, 8, 6, 8, 4, 6, 4, 4, 2, 8, 7, 12, 6, 15, 10, 10, 5, 16, 12, 16, 8, 12, 8, 8, 4, 15, 12, 18, 9, 18, 12, 12, 6, 12, 9, 12, 6, 9, 6, 6, 3, 12, 10, 16, 8, 18, 12, 12, 6, 16, 12
OFFSET
0,2
COMMENTS
For all i, j: A286531(i) = A286531(j) => a(i) = a(j).
FORMULA
a(n) = A000005(A163511(n)).
a(n) = A106737(A054429(n)).
For all n >= 0, a(2^n) = n+2.
PROG
(PARI) A324183(n) = if(!n, 1, n = ((3<<#binary(n\2))-n-1); my(e=0, m=1); while(n>0, if(!(n%2), m *= (1+e); e=0, e++); n >>= 1); (m*(1+e)));
(PARI)
A163511(n) = if(!n, 1, my(p=2, t=1); while(n>1, if(!(n%2), (t*=p), p=nextprime(1+p)); n >>= 1); (t*p));
A324183(n) = numdiv(A163511(n));
(PARI)
A054429(n) = if(!n, n, ((3<<#binary(n\2))-n-1)); \\ After code in A054429
A106737(n) = sum(k=0, n, (binomial(n+k, n-k)*binomial(n, k)) % 2);
(Python)
def A324183(n):
if n:
c = 1
while n:
c *= (s:=(~n&n-1).bit_length()+1)
n >>= s
return c*(s+1)//s
return 1 # Chai Wah Wu, Jul 25 2023
KEYWORD
nonn
AUTHOR
Antti Karttunen, Feb 17 2019
STATUS
approved